Problem 26 - A Graphical Approach to College Algebra

Question

Find all complex solutions for each equation by hand. Do not use a calculator. $$\frac{1}{x+3}+\frac{4}{x+5}=\frac{2}{x^{2}+8 x+15}$$

Step-by-Step Solution

Answer

No valid solutions exist.

Step 1: Simplify the Equation

The equation is \( \frac{1}{x+3} + \frac{4}{x+5} = \frac{2}{x^2 + 8x + 15} \). Notice that the denominator on the right-hand side, \(x^2 + 8x + 15\), can be factored as \((x+3)(x+5)\). Thus, rewrite the equation with a common denominator: \( \frac{1(x+5) + 4(x+3)}{(x+3)(x+5)} = \frac{2}{(x+3)(x+5)} \).

Step 2: Combine Expressions in Numerator

Combine the numerator by expressing \( 1(x+5) + 4(x+3) \) as \( x + 5 + 4x + 12 \). Simplify to get \( 5x + 17 \) as the new numerator. This leads to \( \frac{5x + 17}{(x+3)(x+5)} = \frac{2}{(x+3)(x+5)} \).

Step 3: Cross-multiply to Eliminate Denominators

Cross-multiplying gives \( 5x + 17 = 2 \). This simplifies our problem to a simple linear equation.

Step 4: Solve the Linear Equation

Solve \( 5x + 17 = 2 \) to find \( x \). Subtract 17 from both sides to get \( 5x = -15 \). Then, divide by 5 to isolate \( x \), resulting in \( x = -3 \).

Step 5: Check Solution Validity

Check if \( x = -3 \) is valid by substituting back into the original equation. Since \( x = -3 \) makes both \( x + 3 \) and \( (x+3)(x+5) = 0 \), it is an invalid solution (division by zero). Thus, no other solutions are possible without recalculating.

Key Concepts

Factoring Polynomials

Factoring is a key step in solving polynomial equations and can simplify expressions considerably when dealing with algebra.To factor, you aim to express the polynomial as a product of its factors.For example, the polynomial \( x^2 + 8x + 15 \) can be factored into \((x + 3)(x + 5)\).
This involves finding two numbers that multiply to give the constant term (15 in this case) and add to give the middle coefficient (8).
This results in finding factors like 3 and 5 that satisfy both conditions.Factoring allows us to see the structure of the polynomial and simplifies expressions by reducing the need to handle complex denominators separately.

• It helps in simplifying fractions where the denominator contains a polynomial.
• Factoring can also reveal roots or zeros of the polynomial.
• Facilitates solution through methods like cross-multiplication, as seen in the given problem.

Remember, factoring is not just a mechanical step; it unveils deeper insights into the equation's behavior.

Solving Equations

Solving an equation means finding the value(s) of the variable(s) that make the equation true.In the given exercise, the goal was to find a solution to the equation by simplifying and manipulating it.
The equation was simplified to:\[\frac{5x + 17}{(x+3)(x+5)} = \frac{2}{(x+3)(x+5)}\]This step simplifies the equation to find any potential solutions.A critical aspect of solving equations is the simplification process:

• Expressions in the numerators and denominators might require factoring or expanding.
• The results hinge upon equalizing both sides of the equation.
• Simplified forms should lead to an uncomplicated linear or quadratic equation.

Simplification reveals any existing solution in a straightforward manner, as we saw when the problem pared down to straightforward arithmetic: \(5x + 17 = 2\).

Cross-Multiplication

Cross-multiplication is a technique used to eliminate fractions from an equation, facilitating its solution by transforming it into a simple expression.In the exercise, after factoring and rewriting with a common denominator, cross-multiplying allowed us to proceed with:\[ 5x + 17 = 2 \]This is accomplished by multiplying each denominator with the opposite side’s numerator, thereby removing the fractional aspect of the equation.
Cross-multiplication is particularly useful when:

• Dealing with equations represented as fractions or ratios.
• A cross equality exists where two fractions are set against each other.
• Solving rational equations where direct simplification is complicated.

While it efficiently resolves the fraction aspect, always consider potential undefined solutions, such as division by zero scenarios, as critical checks.
In this problem, cross-multiplying lead to identifying \( x = -3 \) as an apparent but invalid solution due to division by zero.